Consider a functional Q[n] that scales homogeneously, Q[${\mathit{n}}_{\lambda }$]=${\lambda }^{\mathit{k}}$Q[n], where ${\mathit{n}}_{\lambda }$(r)=${\lambda }^3$n(λr). It is already known that Q[n] is related to its functional derivative, q([n];r)==δQ[n]/δn, by kQ[n] =-F${\mathit{d}}^3$r n(r)r⋅∇q([n];r). We here prove that q([n];r) is related to its functional derivative by kq([n];r)=-F${\mathit{d}}^3$r’n(r’)r’⋅∇’{δq([n];r) /δn(r’)}-r⋅∇q([n];r) and that there also exists a homo- geneouslike scaling relation for q([n];r): q([${\mathit{n}}_{\lambda }$];r)=${\lambda }^{\mathit{k}}$q([n];λr). In addition, δq([n];r)/δn(r’)=δq([n];r’)/δn(r) because q([n];r) is the functional derivative of Q[n]. Based upon these exact properties of q([n];r) it is proved that if a trial potential q¯([n];r) satisfies kQ[n]=-F${\mathit{d}}^3$r n(r)r⋅∇q¯([n];r) and δq¯([n];r)/δn(r’)=δq¯([n];r’)/δn(r), then kq([n];r)=-F${\mathit{d}}^3$r’n(r’)r’⋅∇’{δq¯([n]; r)/δn(r’)}-r⋅∇q¯([n];r). If q¯([n];r) further satisfies q¯([${\mathit{n}}_{\lambda }$];r)=${\lambda }^{\mathit{k}}$q([n];λr), we prove that q([n];r)=q¯([n];r), which means that q¯ is exact. Application of the theorem to density-functional theory is discussed.

• Received 8 October 1990

DOI:https://doi.org/10.1103/PhysRevA.44.54